Theory of Equation – Nature of the roots of an equation


Theory of Equation / Monday, October 14th, 2019

Nature of the roots of an equation

Now, we shall have two important theorems on the nature of the roots an algebraic equation. These theorems have wide range of applications in the theory of equations.

 Theorem 01

In an algebraic equation with rational coefficients irrational roots are occur in conjugate pairs.

That is α – √β is another root of the equation f(x) = 0, if α + √β be a root.

 Theorem 02

In an algebraic equation with real coefficients, imaginary roots occur in conjugate pairs.

That is, if α + iβ (β ≠ 0) is a root of the equation f(x) = 0, α – iβ is another root.

Note

The occurrence of irrational or imaginary roots of multiplicity m is an algebraic equation with rational or real coefficients implies the occurrence of conjugate irrational or imaginary roots of the same multiplicity.

General Properties of Algebraic Equations

In this section a few general properties of algebraic equations will be discussed. The properties indicate the existence and location of real roots of an equation.

 Property I

If, for two distinct real numbers α and β, f(α) and f(β) are of opposite signs, then the polynomial equation f(x) = 0 must have one real root lying between α and β.

 Property II

(i) If f(α) and f(β) are of opposite signs, then the equation f(x) = 0 has an odd number of real roots between α and β.

(ii) If f(α) and f(β) are of same sign, then the equation f(x) = 0 has either an even number of real roots or no real root between α and β.

 Property III

(i) An equation of an odd degree must have at least one real root, opposite in sign to that of the last term, the leading term being positive.

(ii) An equation of an odd degree, whose last term is negative, has at least two real roots, one positive and the other negative.

Multiple Roots

Let the algebraic equation f(x) = 0 of degree n have n roots α1 ,  α2 ,  α3 ,  …, αn of which the first r (r < n) roots are equal to one another, each of them being equal to α1 . Then f(x) = (x – α1)rφ(x), where φ(α1) ≠ 0. Under this condition it can be said that α1 is a root of the equation f(x) = 0 of multiplicity r.

 Theorem

If α1 is a root of the equation f(x) = 0 of multiplicity r, then α1 is a root of the equation f’(x) = 0 of multiplicity (r – 1), where f’(x) is the first derived function of f(x).

 Working Rule

In order to obtain the multiple root of the equation f(x) = 0, the highest common factor (H.C.F) of f(x) and f’(x) are worked out. The HCF will give the common root of f(x) = 0 and f’(x) = 0.

If the HCF is of the form (x – α)m, then α is a root of the equation f(x) = 0 of multiplicity (m + 1).

 Question 01

If one root of the equation x4 – 3x3 – 5x2 + 9x – 2 = 0 is 2 – √3, find the other roots.

Solution:

The coefficients of the given equation are rational numbers. Irrational roots in such equation must occur in conjugate pairs.

If 2 – √3 is a root of this equation, 2 + √3 must also be another root.

So,

\[\{x-(2-\sqrt{3})\}\{x-(2+\sqrt{3})\}\]

\[={{(x-2)}^{2}}-{{(\sqrt{3})}^{2}}\]

\[={{x}^{2}}-4x+4-3\]

\[={{x}^{2}}-4x+1\]

should be a factor of (x4 – 3x3 – 5x2 + 9x – 2).

On being divided by this factor, the given equation reduces to

\[{{x}^{2}}+x-2=0\]

\[\Rightarrow (x+2)(x-1)=0\]

\[i.e.,x=-2,1\]

Therefore, the other roots of the equation are: -2, 1, 2 + √3.

 Question 02

Solve x4 – x3 + 2x2 – 2x + 4 = 0, (1 + i) being one of its roots.

Solution:

Since the coefficients of the equation are real and 1 + i is a root of the equation, so 1 – i is also a root of the equation.

So,

\[\{x-(1-i)\}\{x-(1+i)\}\]

\[={{(x-1)}^{2}}-{{(i)}^{2}}\]

\[={{x}^{2}}-2x+1+1\]

\[={{x}^{2}}-2x+2\]

is a factor of (x4 – x3 + 2x2 – 2x + 4).

Let,

\[{{\text{x}}^{\text{4}}}\text{ }{{\text{x}}^{\text{3}}}+\text{ 2}{{\text{x}}^{\text{2}}}\text{ 2x }+\text{ 4}=({{x}^{2}}-2x+2)q(x)\]

\[\therefore q(x)=\frac{{{\text{x}}^{\text{4}}}\text{ }{{\text{x}}^{\text{3}}}+\text{ 2}{{\text{x}}^{\text{2}}}\text{ 2x }+\text{ 4}}{{{x}^{2}}-2x+2}\]

\[\therefore q(x)={{x}^{2}}+2x+2\]

Thus the other roots is given by,

\[{{x}^{2}}+2x+2=0\]

\[i.e.,x=-1\pm i\]

Thus the roots of the given equation are

\[1\pm i,-1\pm i\]

 Question 03

If A, B, …, K and a, b, …,k and l are all reals, show that

\[\frac{{{A}^{2}}}{x-a}+\frac{{{B}^{2}}}{x-b}+…+\frac{{{K}^{2}}}{x-k}=x+l\]

Has all its roots real.

Solution:

If possible, let α + iβ be an imaginary root of the equation.

Then we have,

\[\frac{{{A}^{2}}}{\alpha +i\beta -a}+\frac{{{B}^{2}}}{\alpha +i\beta -b}+…+\frac{{{K}^{2}}}{\alpha +i\beta -k}=\alpha +i\beta +l……(1)\]

Again, since the coefficients of the equations are real, α – iβ is also a root of the equation. Thus we have

\[\frac{{{A}^{2}}}{\alpha -i\beta -a}+\frac{{{B}^{2}}}{\alpha -i\beta -b}+…+\frac{{{K}^{2}}}{\alpha -i\beta -k}=\alpha -i\beta +l……(2)\]

Subtracting (2) from (1), we get

\[{{A}^{2}}\left[ \frac{1}{(\alpha -a)+i\beta }-\frac{1}{(\alpha -a)-i\beta } \right]+{{B}^{2}}\left[ \frac{1}{(\alpha -b)+i\beta }-\frac{1}{(\alpha -b)-i\beta } \right]\]

\[+…+{{K}^{2}}\left[ \frac{1}{(\alpha -k)+i\beta }-\frac{1}{(\alpha -k)-i\beta } \right]=2i\beta \]

\[\Rightarrow {{A}^{2}}\left[ \frac{-2i\beta }{{{(\alpha -a)}^{2}}+{{\beta }^{2}}} \right]+{{B}^{2}}\left[ \frac{-2i\beta }{{{(\alpha -b)}^{2}}+{{\beta }^{2}}} \right]+…+{{K}^{2}}\left[ \frac{-2i\beta }{{{(\alpha -k)}^{2}}+{{\beta }^{2}}} \right]=2i\beta \]

\[\Rightarrow 2i\beta \left[ \frac{{{A}^{2}}}{{{(\alpha -a)}^{2}}+{{\beta }^{2}}}+\frac{{{B}^{2}}}{{{(\alpha -b)}^{2}}+{{\beta }^{2}}}+…+\frac{{{K}^{2}}}{{{(\alpha -k)}^{2}}+{{\beta }^{2}}}+1 \right]=0\]

Since A, B, …, K and a, b, …,k and α, β are all real, the expressions within the brackets cannot be zero. Hence β = 0. Thus there is no imaginary root of the equation and so the roots are all real.

 Question 04

Find the equation of the fourth degree with rational coefficients, one root of which is √2 + √3i.

Solution:

Since one root is √2 + √3i, the other root is √2 – √3i.

So,

\[\{x-(\sqrt{2}+\sqrt{3}i)\}\{x-(\sqrt{2}-\sqrt{3}i)\}\]

\[={{(x-\sqrt{2})}^{2}}-{{(\sqrt{3}i)}^{2}}\]

\[={{x}^{2}}-2\sqrt{2}x+2+3\]

\[={{x}^{2}}-2\sqrt{2}x+5\]

Since the equation is of degree four with rational coefficients. Therefore other quadratic factor must be

\[{{x}^{2}}+2\sqrt{2}x+5\]

Hence the required equation is

\[({{x}^{2}}-2\sqrt{2}x+5)({{x}^{2}}+2\sqrt{2}x+5)=0\]

\[\Rightarrow {{({{x}^{2}}+5)}^{2}}-{{(2\sqrt{2}x)}^{2}}=0\]

\[\Rightarrow {{x}^{4}}+10{{x}^{2}}+25-8{{x}^{2}}=0\]

\[\Rightarrow {{x}^{4}}+2{{x}^{2}}+25=0\]

 Question 05

Show that the equation (x – a)3+ (x – b)3 + (x – c)3 + (x – d)3 = 0 where a, b, c, d are positive and not all equal has only one real root.

Solution:

Let f(x) = (x – a)3+ (x – b)3 + (x – c)3 + (x – d)3 since the equation is of degree 3, it has either only one real root or three real roots.

Let us assume that the equation has more than one real root. Let α, β be such. Then f’(x) = 0 has a real root between α and β.

\[f'(x)=3{{(x-a)}^{2}}+3{{(x-b)}^{2}}+3{{(x-c)}^{2}}+3{{(x-d)}^{2}}\]

\[\Rightarrow f'(x)=12{{x}^{2}}-6x(a+b+c+d)+3({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}})\]

The discriminant of the equation f’(x) = 0 is

\[36{{(a+b+c+d)}^{2}}-144({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}})\]

Since a, b, c, d are all positive and not all equal, so

\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}>\frac{{{(a+b+c+d)}^{2}}}{4}\]

Therefore, the discriminant of the equation f’(x) = 0 is negative and so f’(x) has no real root. This proves that f(x) = 0 has only one real root.

 Question 06

Prove that the equation xn – qxn-m + r = 0 has two equal roots if

\[{{\left\{ \frac{q}{n}(n-m) \right\}}^{n}}={{\left\{ \frac{r}{m}(n-m) \right\}}^{m}}\]

Solution:

Let f(x) = xn – qxn-m + r = 0

Then

\[f'(x)=n{{x}^{n-1}}-q(n-m){{x}^{n-m-1}}\]

If f(x) = 0 has two equal roots then f(x) = 0 and f’(x) = 0 have a common root and the condition is obtained by eliminating x between these two equations.

From f’(x) = 0 we have,

\[n{{x}^{n-1}}-q(n-m){{x}^{n-m-1}}=0\]

\[\Rightarrow {{x}^{m}}.n-q(n-m)=0\]

\[\Rightarrow x={{\left\{ \frac{q(n-m)}{n} \right\}}^{\frac{1}{m}}}\]

Putting this value of x in f(x) = 0, we have

\[{{\left\{ \frac{q(n-m)}{n} \right\}}^{\frac{n}{m}}}-q{{\left\{ \frac{q(n-m)}{n} \right\}}^{\frac{n-m}{m}}}+r=0\]

\[\Rightarrow {{\left\{ \frac{q(n-m)}{n} \right\}}^{\frac{n}{m}}}\left\{ 1-\frac{nq}{q(n-m)} \right\}+r=0\]

\[\Rightarrow {{\left\{ \frac{q(n-m)}{n} \right\}}^{\frac{n}{m}}}=\frac{r(n-m)}{m}\]

\[\therefore {{\left\{ \frac{q}{n}(n-m) \right\}}^{n}}={{\left\{ \frac{r}{m}(n-m) \right\}}^{m}}\]

 Question 07

Show that the equation xn – nqx + (n – 1)r = 0 will have a pair of equal roots if qn = rn-1.

Solution:

Let f(x) = xn – nqx + (n – 1)r

\[\therefore f'(x)=n{{x}^{n-1}}-nq\]

\[\therefore f'(x)=0\Rightarrow x={{q}^{\frac{1}{n-1}}}\]

If f(x) = 0 has a pair of equal roots, then this root must be a root of f’(x) = 0.

Hence q1/(n-1) is a root of f(x) = 0

\[\therefore {{q}^{\frac{n}{n-1}}}-nq.{{q}^{\frac{1}{n-1}}}+(n-1)r=0\]

\[\Rightarrow (1-n){{q}^{\frac{n}{n-1}}}=(1-n)r\]

\[\therefore {{q}^{n}}={{r}^{n-1}}\]

 Question 08

Determine the multiple roots of the equation x5 + 2x4 + 2x3 + 4x2 + x + 2 = 0

Solution:

Let,

\[f(x)={{\text{x}}^{\text{5}}}+\text{ 2}{{\text{x}}^{\text{4}}}+\text{ 2}{{\text{x}}^{\text{3}}}+\text{ 4}{{\text{x}}^{\text{2}}}+\text{ x }+\text{ 2}\]

Then,

\[f'(x)=5{{x}^{4}}+8{{x}^{3}}+6{{x}^{2}}+8x+1\]

The HCF of f(x) and f’(x) is obtained as x2 + 1.

\[{{x}^{2}}+1=(x+i)(x-i)\]

Therefore f(x) = 0 has two multiple roots i of order 2 and –i of order 2.

 Question 09

Solve the equation x4 – x3 + 2×2 – x + 1 = 0 which has four distinct roots of equal moduli.

Solution:

Let r be the modulus. Since the roots are all distinct, two possibilities may occur.

I. Two roots are real and two complexes. The roots are r, -r, r(cosθ ± isinθ)

II. All are complex roots. The roots are r(cosθ ± isinθ) and r(cosφ ± isinφ).

Case I.

The given equation is identical with

\[(x-r)(x+r)({{x}^{2}}-2r\cos \theta x+{{r}^{2}})=0\]

\[\Rightarrow {{x}^{4}}-2r\cos \theta {{x}^{3}}+2{{r}^{3}}\cos \theta x-{{r}^{4}}=0\]

This cannot happen, since there is a term containing x2 in the given equation.

Case II.

The given equation in identical with

\[({{x}^{2}}-2r\cos \theta x+{{r}^{2}})({{x}^{2}}-2r\cos \phi x+{{r}^{2}})=0\]

\[\Rightarrow {{x}^{4}}-2r(\cos \theta +\cos \phi ){{x}^{3}}+2{{r}^{2}}(1+2\cos \theta \cos \phi ){{x}^{2}}\]

\[-2{{r}^{3}}(\cos \theta +\cos \phi )x+{{r}^{4}}=0\]

\[\therefore 2r(\cos \theta +\cos \phi )=1,~2{{r}^{2}}(1+2\cos \theta \cos \phi )=2,\]

\[2{{r}^{3}}(\cos \theta +\cos \phi ),{{r}^{4}}=1\]

\[\text{This gives r }=\text{ 1},\cos \theta +\cos \phi =\frac{1}{2},2\cos \theta \cos \phi =0\]

\[\therefore ~~either~~\cos \theta =0~~~~or,~~\cos \phi =0\]

\[\text{Taking}~~\text{cos}\theta \text{ }=\text{ }0\text{ we have}~~\cos \phi =\frac{1}{2},~~\sin \phi =\frac{\sqrt{3}}{2},~~\sin \theta =1.\]

\[\text{Taking}~~\text{cos}\phi \text{ }=\text{ }0\text{ we have}~~\cos \theta =\frac{1}{2},~~\sin \theta =\frac{\sqrt{3}}{2},~~\sin \phi =1.\]

\[\text{Therefore the roots are}~~\pm i,~~\frac{1\pm \sqrt{3}i}{2}\]

 Question 10

Show that the equation x3 + 2x2 – 2x – 1 = 0 has one positive root and two negative roots – one lying between -3 and -1 and another lying between -1 and 0.

Solution:

Let,

\[f(x)={{\text{x}}^{\text{3}}}+\text{ 2}{{\text{x}}^{\text{2}}}\text{ 2x }\text{ 1}\]

\[\text{Then}~f(\infty )=+ve,f(0)=-ve\]

Therefore there is at least one positive root.

\[f(0)=-ve,f(-1)=+ve\]

Thus there is at least one root lying between -1 and 0.

\[f(-3)=-ve,f(-1)=+ve\]

Hence there is at least one root lying between -3 and -1.

Since the equation has exactly three roots, only one is positive, only one lies between -1 and 0 and only one lies between -3 and -1.

 

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